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How many moles of solid Ba(NO3)2 should be added to 300 ml of .2 M Fe(NO3)3 to increase the concentration of NO3- ion to 1 M?

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    0.300 L x 0.2 M = mols Fe(NO3)3
    (NO3^-) = 3 times that = 0.18 mols NO3^-present.

    (NO3^-) must be what to make it 1 M. M x L = mols; mols = 1M x 0.300 L = 0.3 mols needed.

    We have 0.18 mols so we must add 0.300 - 0.180 = 0.120 mols.

    Since Ba(NO3)2 contains 2 mols NO3^-/ mol, then 0.120/2 = 0.060 mols Ba(NO3)2 we must add.

    Now we check it to see if we are ok.
    mols NO3^- in 300 mL of 0.2 M Fe(NO3)3 = 0.3 L x 0.2 M x (3 mols NO3^-/1 mol Fe(NO3)3 = 0.18 mols present.

    Adding 0.06 mols Ba(NO3)2 gives us
    0.06 mols Ba(NO3)2 x (2 mols NO3^-/1 mol Ba(NO3)2 = 0.120 mols NO3^-.

    0.180 mols NO3^- present at beginning + 0.120 mols NO3^- added with barium nitrate = 0.300 mols NO3^- total.

    Molarity = # mols/L = 0.300 mols/0.300 L = 1 M.

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